$f(x, y) = 3x + ye^y$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $e^y + ye^y$ (Choice B) B $3$ (Choice C) C $0$ (Choice D) D $3 + e^y + ye^y$
Solution: Taking a partial derivative with respect to $x$ means treating $y$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial x} &= \dfrac{\partial}{\partial x} \left[ 3{x} + ye^y\right] \\\\ &= \dfrac{\partial}{\partial x} \left[ 3{x} \right] + \dfrac{\partial}{\partial x} \left[ ye^y\right] \\\\ &= 3+0 \end{aligned}$ Note that $\dfrac{\partial}{\partial x} \left[ ye^y \right] = 0$ since we're treating $y$ as a constant. In conclusion, $\dfrac{\partial f}{\partial x}=3$